# Tadpole diagrams in $\phi^3$ theory

In "Quantum Field Theory" by Mark Srednicki, Chapter 9 page 67, after he proves that $\langle 0|\phi(x)|0 \rangle$ vanishes (meaning sum of all connected diagrams with a single source is zero), he makes the following claim:

Consider now that same infinite set of diagrams, but replace the single source in each of them with some other subdiagram. Here is the point: no matter what this replacement subdiagram is, the sum of all these diagrams is still zero. Therefore we need not bother to compute any of them! The rule is this: ignore any diagram that, when a single line is cut, falls into two parts, one of which has no sources. All of these diagrams (known as tadpoles) are canceled by the $Y$ counterterm, no matter what subdiagram they are attached to.

The most important question that I am wondering is how he arrived at this conclusion from his proof that the $Y$ counterterm can be used to make $\langle0|\phi(x)|0\rangle$ zero.

Also, what does he mean by "subdiagram" here? A part of the diagram formed by cutting one of the diagram that has a source, or a part of any diagram that does not necessarily have a source? Is he replacing each of the different diagrams with single source with identical subdiagram or replacing each source with different subdiagrams? (Since "subdiagram" is singular, I am guessing they are all replaced with identical subdiagrams.)

1. Ref.1 is considering $\varphi^3$ theory $$\tag{1} {\cal L}(J)~=~\frac{1}{2}Z_{\varphi}\partial^{\mu}\varphi\partial_{\mu}\varphi - \frac{1}{2}Z_{m}m^2\varphi^2 - \frac{1}{6}Z_{g}g\varphi^3+(Y+J)\varphi.$$ To read the Feynman diagrams in Ref. 1, note that the source $J(x)$ is drawn as a black bullet $\bullet$, and the counterterm $Y(x)$ is drawn as a cross $\times$.
2. Technically, $$\tag{2} \langle 0 | \varphi(x) | 0 \rangle~=~ \frac{1}{i} \left. \frac{\delta W(J)}{\delta J(x)}\right|_{J=0}$$ is the sum of all connected Feynman diagrams with a single source $J(x)$, with the source removed/striped.
3. We have adjusted the $Y$ counterterm in the $\varphi^3$ theory, so that the sum (2) is zero: $$\tag{3} \langle 0 | \varphi(x) | 0 \rangle~=~0.$$
4. We now consider the same collection of Feynman diagrams, with the single source $J(x)$ replaced by a fixed but arbitrary subdiagram $SD(x)$. The corresponding sum will then again vanish $$\tag{4} \int\!d^4x~ SD(x)\langle 0 | \varphi(x) | 0 \rangle~=~0,$$ due eq. (3) and the distributive law.